Aptitude - Numbers

  35423         317 x 89 = 317 x (90 -1 )
+  7164                  = (317 x 90 - 317)
+ 41720                  = (28530 - 317)
  -----                  = 28213
  84307
- 28213
  -----
  56094
  -----

For every natural number n, (xⁿ - aⁿ) is completely divisible by (x - a).

45 = 5 x 9, where 5 and 9 are co-primes.

Unit digit must be 0 or 5 and sum of digits must be divisible by 9.

Among given numbers, such number is 202860.

(3²⁵ + 3²⁶ + 3²⁷ + 3²⁸) = 3²⁵ x (1 + 3 + 3² + 3³) = 3²⁵ x 40

     = 3²⁴ x 3 x 4 x 10

     = (3²⁴ x 4 x 30), which is divisible by30.

Let x = 6q + 3.

Then, x² = (6q + 3)²

   = 36q² + 36q + 9

   = 6(6q² + 6q + 1) + 3

Thus, when x² is divided by 6, then remainder = 3.

Let the numbers be a and b. Then, a + b = 12 and ab = 35.

	a + b	=	12			1	+	1		=	12
	ab		35			b		a			35

Sum of reciprocals of given numbers =	12
	35

When n is even. (xⁿ - aⁿ) is completely divisibly by (x + a)

(17²⁰⁰ - 1²⁰⁰) is completely divisible by (17 + 1), i.e., 18.

   (17²⁰⁰ - 1) is completely divisible by 18.

   On dividing 17²⁰⁰ by 18, we get 1 as remainder.

We know that (12 + 22 + 32 + ... + n2) =	1	n(n + 1)(2n + 1)
	6

Putting n = 10, required sum =		1	x 10 x 11 x 21		= 385
		6